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Homogeneous system

  1. Oct 20, 2009 #1

    In a book I'm reading about linear algebra it's mentioned that in order for the homogeneous system
    Ax = 0
    to have a solution (other than the trivial solution) the coefficient Matrix must be singular.
    The thing is, I can't remember (the wikipedia page on homogeneous systems didn't turn up anything) why if A is invertible, then the system does not have non-zero solutions.

    Any help on why this is so is appreciated.

    Thank you in advance
    Last edited: Oct 20, 2009
  2. jcsd
  3. Oct 20, 2009 #2


    Staff: Mentor

    If A is invertible, then A-1 exists.
    Ax = 0 ==> A-1Ax = A-10 = 0 ==> x = 0

    Another way to look at it is that, for A-1 to exist, both it and A must be one-to-one. The equation Ax = 0 obviously has at least one solution, x = 0, but the one-to-oneness prevents it from having any additional solutions.
  4. Oct 21, 2009 #3
    That makes sense, thank you very much for the response Mark44.
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